【Derivation】MarkDown Letex编码 之 正态分布特征函数证明

**求证：$\varphi（u）=e^{jau-\frac{1}{2}u^2\sigma^2} \ \ \ , t\in R $**  **证：**  * *   $$\varphi（u）=\int _ {-\infty} ^ {+\infty} e^{jux}f(x)dx$$   $$=\int_ {-\infty}^{+\infty} e^{jux} \frac{1}{\sqrt{2\pi\sigma^2}}  e^{- \frac{(x-a)^2}{2\sigma^2}}dx$$* 整理，得：* $$\varphi（u）= \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} e^{jux}   e^{- \frac{(x-a)^2}{2\sigma^2}}dx $$* * beacuse $|jx  e^{jux}   e^{- \frac{(x-a)^2}{2\sigma^2}}| \leq |x| e^{jux}   e^{- \frac{(x-a)^2}{2\sigma^2}}$ and $ \frac{1}{\sqrt{2\pi}}|x| e^{jux}   e^{- \frac{(x-a)^2}{2\sigma^2}} < +\infty$ , $so $可以对$\varphi（u）$求$u$的一阶导数，* 有：  $$\varphi \prime（u）= \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} {jx}\ e^{jux}   e^{- \frac{(x-a)^2}{2\sigma^2}}dx   $$综合可推： $$j{(u-j\frac{a}{\sigma^2})\varphi (u)}+\frac{j{\varphi \prime(u) } } {\sigma^2} $$$$=$$ $$ \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} { ( ju-\frac{x-a}{\sigma^2} })\ e^{jux}   e^{- \frac{(x-a)^2}{2\sigma^2}}dx  =$$$$  \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} { ( ju-\frac{x-a}{\sigma^2} })\ e^{jux- \frac{(x-a)^2}{2\sigma^2}}dx $$ $$=\frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} 1de^{jux- \frac{(x-a)^2}{2\sigma^2}} $$$$=\frac{1}{\sqrt{2\pi\sigma^2}}[e^{jux- \frac{(x-a)^2}{2\sigma^2}}]|_{-\infty}^{+\infty}=0$$即得微分方程$${u\varphi (u)-j\frac{a}{\sigma^2}\varphi (u)}+\frac{{\varphi \prime(u) } } {\sigma^2}=$$$${(u\sigma^2 -ja)}{\varphi (u)}+{\varphi \prime(u) } =0$$即得微分方程++++分水岭，从后往前推+++++++$${\varphi (u)}+\frac{{\varphi \prime(u) } } {u\sigma^2 -ja} $$$$={u\varphi (u)}+\frac{{\varphi \prime(u) } } {\sigma^2 -\frac{ja}{u}}=0 $$求解：$$\frac{\varphi\prime(u)}{\varphi(u)}=-u\sigma^2+ja$$解得：$$\ln\varphi (u)=-\frac{1}{2}u^2D(x)+jau+C$$进一步化简：$$\varphi (u)=e^Ce^{-\frac{1}{2}u^2D(x)+jau}$$令$u=0,e^C=\varphi(0)=E[E^(j0X)]=E[e^0]=1$,故$C=0;$代入通解为：$$\varphi (u)=e^{jau-\frac{1}{2}u^2D(x)}$$由以上推导，**正态分布特征函数表达式** 得证

**求证：$\varphi（u）=e^{jau-\frac{1}{2}u^2\sigma^2} \ \ \ , t\in R $**  **证：**  * *   $$\varphi（u）=\int _ {-\infty} ^ {+\infty} e^{jux}f(x)dx$$   $$=\int_ {-\infty}^{+\infty} e^{jux} \frac{1}{\sqrt{2\pi\sigma^2}}  e^{- \frac{(x-a)^2}{2\sigma^2}}dx$$* 整理，得：* $$\varphi（u）= \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} e^{jux}   e^{- \frac{(x-a)^2}{2\sigma^2}}dx $$* * beacuse $|jx  e^{jux}   e^{- \frac{(x-a)^2}{2\sigma^2}}| \leq |x| e^{jux}   e^{- \frac{(x-a)^2}{2\sigma^2}}$ and $ \frac{1}{\sqrt{2\pi}}|x| e^{jux}   e^{- \frac{(x-a)^2}{2\sigma^2}} < +\infty$ , $so $可以对$\varphi（u）$求$u$的一阶导数，* 有：  $$\varphi \prime（u）= \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} {jx}\ e^{jux}   e^{- \frac{(x-a)^2}{2\sigma^2}}dx   $$综合可推： $$j{(u-j\frac{a}{\sigma^2})\varphi (u)}+\frac{j{\varphi \prime(u) } } {\sigma^2} $$$$=$$ $$ \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} { ( ju-\frac{x-a}{\sigma^2} })\ e^{jux}   e^{- \frac{(x-a)^2}{2\sigma^2}}dx  =$$$$  \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} { ( ju-\frac{x-a}{\sigma^2} })\ e^{jux- \frac{(x-a)^2}{2\sigma^2}}dx $$ $$=\frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} 1de^{jux- \frac{(x-a)^2}{2\sigma^2}} $$$$=\frac{1}{\sqrt{2\pi\sigma^2}}[e^{jux- \frac{(x-a)^2}{2\sigma^2}}]|_{-\infty}^{+\infty}=0$$即得微分方程$${u\varphi (u)-j\frac{a}{\sigma^2}\varphi (u)}+\frac{{\varphi \prime(u) } } {\sigma^2}=$$$${(u\sigma^2 -ja)}{\varphi (u)}+{\varphi \prime(u) } =0$$即得微分方程++++分水岭，从后往前推+++++++$${\varphi (u)}+\frac{{\varphi \prime(u) } } {u\sigma^2 -ja} $$$$={u\varphi (u)}+\frac{{\varphi \prime(u) } } {\sigma^2 -\frac{ja}{u}}=0 $$求解：$$\frac{\varphi\prime(u)}{\varphi(u)}=-u\sigma^2+ja$$解得：$$\ln\varphi (u)=-\frac{1}{2}u^2D(x)+jau+C$$进一步化简：$$\varphi (u)=e^Ce^{-\frac{1}{2}u^2D(x)+jau}$$令$u=0,e^C=\varphi(0)=E[E^(j0X)]=E[e^0]=1$,故$C=0;$代入通解为：$$\varphi (u)=e^{jau-\frac{1}{2}u^2D(x)}$$由以上推导，**正态分布特征函数表达式** 得证